英语翻译Of course,this result is well-known now,but in the 1950’

英语翻译
Of course,this result is well-known now,but in the 1950’s this topic was only beginning
to surface.We color the edges of the complete graph with 6 vertices (and so 15 edges) with two colors.Showwe can always find three points such that the edges joining them are of the same color.The proof is quite simple.Take any vertex,name it a.It has 5 edges,so by the pigeonhole principle at least three of those are of the same color.Say ab,ac and ad are of the same color,blue (without loss of generality).Then if either bc,bd or cd are blue,we are done.But if none is,then bcd is a red triangle.Either way,we have found three points in the complete graph of three vertices such that the three edges joining them are of the same color.Remarks.A very similar problem in graph theory,also easily solved using the pigeonhole principle,is the following :suppose there are n people at a party,then show at least two of them have the same number of acquaintances,where a person is not acquainted with himself or herself.This is easy,for the maximum number of acquaintances for one person is n-1.For everyone to have a different number of friends would mean someone at the party has no friends at all,a contradiction to the fact that someone is friends with everyone (let alone that this person would not be at the party if he or she did not know anyone)!Also,consider the complete graph of six vertices,assume all edges have different length.Some edges form triangles between themselves.Prove there is at least one edge that is both the shortest edge of a triangle and the longest edge of another triangle.To prove this,let us start by coloring blue all edges that are the shortest edge of at least one triangle,and coloring the rest,if any,red.Then,we know there is at least one triangle such that its edges are all blue (we cannot have a triangle all red since its shortest edge should be blue).Take the longest edge of that triangle; clearly it is blue because it is the shortest edge of some other triangle.

Of course, this result is well-known now, but in the 1950’s this topic was only beginning to surface.当然,这个结果现在众所周知,但在1950年这个话题仅仅是表面的.
We color the edges of the complete graph with 6 vertices (and so 15 edges) with two colors. 我们用6个顶点（15边）两种颜色来为这个图表着色.
Show we can always find three points such that the edges joining them are of the same color.所以我们总是能够找到三个点它们的边是用相同颜色连接.
The proof is quite simple. Take any vertex, name it a. It has 5 edges, so by the pigeonhole principle at least three of those are of the same color.
样张很简单,任意一个顶点,称作a,它有五条边,根据鸽棚规则它们中的三条都有同样颜色的边
Say ab, ac and ad are of the same color, blue (without loss of generality). Then if either bc, bd or cd are blue, we are done.
比如ab,ac和ad是同样的颜色——蓝色（在一般情况下）.那么bc,bd和cd也是蓝色,我们就完成了.
But if none is, then bcd is a red triangle. Either way, we have found three points in the complete graph of three vertices such that the three edges joining them are of the same color.Remarks.
但如果没有一个是这样,那么bcd是一个红色的三角形.任何一种方式,我们发现三个顶点的整个图表的三点是用相同颜色标识连在一起.
A very similar problem in graph theory, also easily solved using the pigeonhole principle, is the following : suppose there are n people at a party, then show at least two of them have the same number of acquaintances, where a person is not acquainted with himself or herself.
一个和图表理论非常相近的问题,用鸽篷理论也能轻易解决.如下列所述,假如宴会上有n个人,找出至少两个有同等数量熟人的,其中有一个和他或者她不认识.
This is easy, for the maximum number of acquaintances for one person is n-1. For everyone to have a different number of friends would mean someone at the party has no friends at all, a contradiction to the fact that someone is friends with everyone (let alone that this person would not be at the party if he or she did not know anyone)! 这非常简单,因为每个熟人的最大数是n-1.宴会里有不同数量的朋友的这个人意味着没有朋友.与事实矛盾的是一些人是每个人的朋友（如果一个人不认识任何人他会孤立）
Also, consider the complete graph of six vertices,assume all edges have different length. 同时,考虑六个至高点的图表,假设所有的边长度不同.
Some edges form triangles between themselves.Prove there is at least one edge that is both the shortest edge of a triangle and the longest edge of another triangle.来自三角形的一些表证明至少有一条边是另一个三角形最短或者最长的边.
To prove this, let us start by coloring blue all edges that are the shortest edge of at least one triangle, and coloring the rest, if any, red.
为证明这个,让我们至少为一个三角形的所有边着成蓝色,其他任何的边着成红色.
Then, we know there is at least one triangle such that its edges are all blue (we cannot have a triangle all red since its shortest edge should be blue). Take the longest edge of that triangle; clearly it is blue because it is the shortest edge of some other triangle.
那么,我们知道至少有一个三角形的所有边都是蓝色（我们不可能有个所有边是红色的三角形因为最短的边必须是蓝色）,找出那个三角形的最长的边；很明显是蓝色因为它是其它三角形最短的边.
终于翻译完了,可能有不尽人意的地方,望采纳!谢谢!